Formulae [3]
\[sineA=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\]
\[cosineA=\frac{\mathrm{Base}}{\text{Hypotenuse}}\]
\[tangentA=\frac{\text{Perpendicular}}{\mathrm{Base}}\]
\[cotangent A = \frac{\text{Base}}{\text{Perpendicular}}\]
\[secantA=\frac{\text{Hypotenuse}}{\mathrm{Base}}\]
\[cosecantA=\frac{\text{Hypotenuse}}{\text{Perpendicular}}\]
\[\sin\mathrm{A}=\frac{1}{\mathrm{cosec~A}}\quad\mathrm{and}\quad\mathrm{cosec~A}=\frac{1}{\sin\mathrm{A}}\]
\[\cos\mathrm{A}=\frac{1}{\sec\mathrm{A}}\quad\mathrm{and}\quad\mathrm{sec}\mathrm{A}=\frac{1}{\cos\mathrm{A}}\]
\[\tan\mathrm{A}=\frac{1}{\cot\mathrm{A}}\quad\mathrm{and}\quad\cot\mathrm{A}=\frac{1}{\tan\mathrm{A}}\]
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sinθ⋅cosecθ = 1
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cosθ⋅secθ = 1
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tanθ⋅cotθ = 1
\[tanA=\frac{\sin A}{\cos A}\]
\[cotA=\frac{\cos A}{\sin A}\]
Theorems and Laws [1]
If `tan θ = a/b` then prove that `((a sin theta - b cos theta)/(a sin theta + b cos theta)) = ((a^2 - b^2))/((a^2 + b^2))`.
Let `(a sin theta - b cos theta)/(a sin theta + b cos theta)`
Divide both Nr and Dr with cos θ of (a)
`((a sin theta - b cos theta)/cos theta)/((a sin theta + b cos theta)/cos theta)`
`= (tan theta - b)/(a tan theta + b)`
`=(a xx (a/b) - b)/(a xx (a/b) + b)`
`= (a^2 - b^2)/(a^2 + b^2)`
Key Points
For an acute angle A in a right-angled triangle:
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Hypotenuse is the side opposite the right angle.
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Perpendicular is the side opposite angle A.
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Base is the side adjacent to angle A.
